45 Answers
45

Here is the proof of the equivalent statement "Every complex non-constant polynomial $p$ is surjective".
1) Let C be the finite set of critical points ($p'(z)=0$). C is finite by elementary algebra.

2) Remove from the codomain $p(C)$ (and call the resulting open set B) and from the domain its inverse image (again finite) (and call the resulting open set A).

3) Now you get an open map from A to B, which is also closed, because any polynomial is proper (inverse images of compact sets are compact). But B is connected and so $p$ is surjective.

I like this proof because you can try it for real polynomials and it breaks down at step 3) because if you remove a single point from the line you disconnect it, while you can remove a finite set from a plane leaving it connected.

Alternatively presented: Consider the map from the Riemann sphere to the Riemann sphere induced by a polynomial. This is a continuous map from a compact space to a Hausdorff space, and thus its image is closed. It is also trivially holomorphic, and thus, if non-constant, its image is open. But the domain is inhabited and the codomain is connected, so this map must be surjective.
–
Sridhar RameshOct 14 '11 at 20:20

Here is a standard algebraic proof. It suffices to show that if L/$\mathbb{C}$ is a finite extension, then $L=\mathbb{C}$. By passing to a normal closure we assume that $L/\mathbb{R}$ is Galois with Galois group $G$. Let $H$ be the Sylow-2 subgroup of $G$ and $M=L^H$.

By the Fundamental Theorem of Galois Theory, $M/\mathbb{R}$ has odd degree. Let $\alpha\in M$ and $f(x)$ be its minimal polynomial. Then $f(x)$ has odd degree and by the Intermediate Value Theorem, a real root. As $f(x)$ is irreducible, it must have degree one. Then $\alpha\in\mathbb{R}$ and $M=\mathbb{R}$. So $G=H$ is a 2-group. Then $G_1:=Gal(L/\mathbb{C})$ is a 2-group as well.

Assuming that $G_1$ is not trivial, there must exist a degree 2 subextension $K$ of $\mathbb{C}$. But every quadratic complex polynomial has a root (by the quadratic formula), so we have a contradiction.

This proof can be extended to real closed fields, that is ordered fields in which every polynomial of odd degree has at least one root and every positive element is a square (and so has a square root).
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lhfJan 4 '10 at 0:10

3

This prove can also be extended to prove that the "field" that you obtain by adjoining a square root of -1 to the "field" of surreal numbers is algebraically closed. ("field" is not actually precise, because these sets are proper classes)
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Andrea FerrettiFeb 22 '10 at 14:18

2

I learned this proof from Serge Lang in 1965 when I was a freshman in college. He attributed it to Emil Artin (and he was the keeper of the Artin flame, so I'd definitely believe it).
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Victor MillerApr 15 '12 at 2:39

You can prove it using only basic facts about continuity/compactness and the same estimate which makes the winding number/fundamental group of $S^1$ proofs work: first check that if $p(z)=z^n + a_{n-1}z^{n-1} + \ldots + a_0$ is a polynomial then $|p(z)|$ tends to infinity as $|z|$ tends to infinity (this is the "leading term dominates" estimate for large $|z|$). It follows easily that $|p|$ attains a minimum value, since outside a large disc centered at $0$ the value of $|p|$ is really big, and the disc is compact, so $|p|$ attains a minimum on it. We want this minimum value to be zero, so suppose for the sake of contradiction it isn't, then we can change coordinates if necessary so that minimum is attained at $0$, and rescale $p$ so the minimum is $1$.

Then you just have to show that if $p(z) = 1+b_kz^k + \ldots + b_n z^n$ (where $k \geq 1$) then you can make $|p|$ smaller than $1$ for some nonzero $z$. But this is just the same kind of estimate as before: the term $b_kz^k$ dominates the other terms for $z$ small, and we can easily arrange for $b_kz^k$ to be a negative real giving the required contradiction.

The proof has the advantage that it makes the theorem "obvious" once you have some notion of compactness in the plane, so you could use this proof pretty early in a course that talks about functions on $\mathbb R^n$ or $\mathbb C$. As I said before though, what makes the proof tick is the same as what makes the $\pi_1(S^1)$ proofs work, it just uses simpler techniques to get a contradiction. Unfortunately I have no idea who it's due to (which is why I explained it rather than giving a reference...)

This proof did not come from that jstor reference (to an article of Fefferman). It is basically the proof by d'Alambert from 1746 and it is essentially the first correct proof, except that properties of compactness and its consequences (continuous real-valued functions assume a minimum value) were not rigorously established back then.
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KConradFeb 12 '10 at 23:07

3

The result that the minimum of $|p|$, if it exists, must be zero is sometimes called d'Alembert's lemma. But his proof is very hard to follow, because he attempts to solve the equation $p(z)=w$ as a fractional power series in $w$. A much easier proof was given by Argand in 1806, using simple algebra and the geometric interpretation of complex numbers.
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John StillwellMay 17 '10 at 1:07

The proof of the ring structure of $H^* \mathbb CP^n$ is the core connectivity/analytical argument comparable to the intermediate value theorem or $\pi_1 S^1 \simeq \mathbb Z$ (used in most of the other proofs).
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Ryan BudneyJan 4 '10 at 7:08

Here's an extract of my post to sci.math.research from 2001. The proof definitely
uses a "sledgehammer method", but perhaps it has some pedagogical value. I have no
doubt that other people may have come up with similar arguments.

Sketch: It suffices to check that any complex monic polynomial has
a root. Any such polynomial is the characteristic polynomial of
some matrix (one can use the so called companion matrix).
Thus one is reduced to showing that an nxn complex matrix A has
an eigenvalue or equivalently an eigenvector. A may be assumed
to be invertible, since otherwise 0 is an eigenvalue. Then
A acts on complex projective space $P = \mathbb{C}\mathbb{P}^{n-1}$ by sending the
span of v to the span of Av. An eigenvector corresponds to
a fixed point under this action. Since the general linear group
of $\mathbb{C}$ is connected, A can be connected by a path to the identity I
(this can be done explicitly by writing A as a product of elementary
matrices and deforming these to I in the obvious way). It follows
that the A is homotopic to I, and therefore its Lefschetz number on
P coincides with the Euler characteristic of P which is nonzero.
Therefore, A has a fixed point on P.

Added: I should probably have pointed that the conclusion follows from the Lefschetz
fixed point theorem, which was the sledgehammer that I was alluding to.

A recent and very important contribution to the literature on the fundamental theorem of algebra is Joe Shipman's article "Improving the Fundamental Theorem of Algebra," Math. Intelligencer 29 (2007), 9-14, doi:10.1007/BF02986170. Here is one of his results: A field with the property that every polynomial whose degree is a prime number has a root is algebraically closed. This result is sharp in the sense that if any prime is omitted then the conclusion is false.

Shipman's paper should go a long way towards addressing Andrew L's question of whether there is a "purely algebraic proof" of the FTA. The above result of Shipman's shows that we can limit the topology/analysis to proving that every polynomial over $\mathbb{C}$ of prime degree has a root; the rest is pure algebra. If you wanted to try to limit the use of topology or analysis even further, then this part of the proof is where you should focus your attention.

Two (or three) more complex analysis approaches, the first is "essentially the same" as the proof in Alfors I think, but the second is different (I'm afraid I don't have a reference, but I can type up the full proofs if you want):

Let p be a polynomial and n be its degree.

apply the residue theorem to 1/[zp(z)]. If p has no roots, then this function is analytic except at z=0, where it has a simple pole of non-zero residue. But 1/[zp(z)] is bounded, so integrating it along a circular contour centred at the origin gives something inversely proportional to the length of this contour ie this integral can be made as small as possible, which contradicts the residue at z=0 being non-zero.

a variant of the above: If p has no roots, then Cauchy's integral formula implies $$\int_{|z|=r}\frac{dz}{zp(z)}=\frac{2\pi i}{p(0)}\ne0.$$ Now let $r\to\infty$: The integral on the left vanishes in the limit, and we have a contradiction. See Anton R. Schep: A simple complex analysis and an advanced calculus proof of the fundamental theorem of algebra, Amer. Math. Monthly 116 (2009) 67–68.

by Rouche's theorem, p(z) and $z^n$ have the same number of roots inside a large circular contour centred at the origin, and this number is n.

One can rephrase the second argument as follows: if P has no roots, then 1/P is entire, and decays to zero at infinity, hence is zero by Liouville's theorem, a contradiction. Though this proof is perhaps somewhat unsatisfying since Liouville's theorem is generally regarded as being less intuitive than the fundamental theorem of algebra.
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Terry TaoSep 19 '10 at 22:43

Gauss's first proof goes more or less like this. Let $p(z)$ be a polynomial of degree $n$ and complex coefficients. Write $p(x+iy) = a(x,y) + ib(x,y)$, where $a,b$ have real coefficients. The crucial observation is that the branches of $a=0$ and $b=0$ as real curves interlace at infinity (as can be seen from the degree $n$ terms). Also, real algebraic plane curves don't just stop somewhere in the affine plane, so a branch of $a=0$ must be connected to another branch of $a=0$. Now, in between them, there is a branch of $b=0$ which connects to another branch of $b=0$. If the connections alternate, they have to meet and we get a common zero of $a$ and $b$, which gives a complex zero of $p$. If they don't alternate, find a branch of $a=0$ in between the two connecting branches of $b=0$ and repeat.

I think this proof really needs the Jordan curve theorem to be fully justified, which is a bit of an anachronism.

complex-analytic, by referring to number of zeroes = number of poles or Stokes theorem;

using topology to prove that holomorphic maps between compact surfaces are constant or surjectivet;

similarly, but using inverse theorem from calculus;

finding a minimum of $|p(z)|$;

applying the Liouville's theorem to $1/p(z)$;

applying Lefschetz theorem to $\mathbb C\mathbb P^n$ to prove that every linear operator over $\mathbb C$ has eigenvector;

using Galois theory + simple facts about $\mathbb C$ to show that $\mathbb C$ has no algebraic extensions;

as well as extensive historical notes.

Though I don't know whether a translation exists, I think that this collection of articles deserves it and I'm sure the authors will be happy to give their permission to republish. A translation could be done as a project for an undergraduate student with knowledge of the language.

In other words, this map sends a set of roots to the polynomial which has precisely those roots. It suffices for us to show that $\phi$ is surjective.

If the points $[\alpha_i:\beta_i]$ are distinct, it is easy to check that the differential $d\phi$ is nonzero. Hence the polynomial $T^n - 1$ (for example) is a regular value of $\phi$ with exactly $n!$ preimages (here we've used the fact that polynomials factor uniquely into irreducibles). Thus the map $\phi$ has positive degree.

It is a fact that any map of positive degree between compact connected complex manifolds of the same dimension is surjective. (Proof: Since any such map is orientation preserving, the number of preimages of any regular value must be exactly equal to the degree - not just up to multiplicity. Hence the image contains the set of regular values, which is dense by Sard's theorem. But the image is also closed since it is the image of a compact set, hence the map is surjective.)

We conclude that $\phi$ is surjective. In other words, every polynomial of degree $n$ has a factorization into linear factors.

Here is another short proof, showing that no finite extension of $\mathbb{C}$ exists. If $A$ were such an extension of dimension $d$,
then the projective space $\mathbb{P} A \cong \mathbb{CP}^{d-1}$
were a compact commutative Lie group, hence a torus.
There are plenty of ways showing that $\mathbb{CP}^{m}$ is not a torus (if $m>0$). Take your favorite one, and the proof is complete.

It is based on the observation that an irreducible polynomial of degree $n$ on a field $F$ gives rise to a field structure on $F^n$, compatible with the vector addition (just quotient out polynomials with coefficients in $F$ by the maximal ideal generated by the irreducible).
Then the FTA is a consequence of the non-existence of real commutative division algebras in dimension greater than $2$.

This last assertion follows from the observation that in such a $R^n$ you can define an $exp$ which gives an epimorphism $R^n\rightarrow R^n$ \ $0$, and hence a homeo $S^j \times R^k\cong R^n$ \ $0$, which is forbidden by a fundamental group computation (this was in Andrea Ferretti's notes).

In fact you can use this argument to prove that any real irreducible polynomial has degree $1$ or $2$.

I don't think they are the same, you can say that they are related, just like nearly all complex analytic proofs are related. At any rate, this one is much clearer to me (a Lie theorist, not an algebraic topologist) - I can actually digest it at a glance, and it avoids unnecessary cohomological machinery.
–
Victor ProtsakMay 18 '10 at 6:46

to locate an eigenvalue $\lambda$, which is then a root of the monic polynomial $z^n + a_{n-1} z^{n-1} + \ldots + a_0$.

Of course, this argument is usually circular, because most of the standard proofs of the spectral theorem for matrices requires the fundamental theorem of algebra (either by explicitly citing that theorem, or implicitly, by borrowing one of the proofs given here, e.g. by applying Liouville's theorem to the resolvent $(A-zI)^{-1}$) in the first place...

However, one could imagine a weird proof of the spectral theorem that somehow avoids the fundamental theorem and would thus give a non-circular proof of that theorem. I thought about proceeding by showing that the set of diagonalisable matrices is a generic subset of the set of all matrices, but I realised that in order to have enough algebraic geometry to talk about "generic", I need to know the ambient field is algebraically closed, which of course is precisely the fundamental theorem of algebra. Deducing the spectral theorem for matrices from the spectral theory of more general objects, such as elements of a C^* algebra, doesn't work either, for much the same reason.

Perhaps it is best to view the above arguments not as proofs of the fundamental theorem of algebra, but rather as "consistency checks" that show that this result is compatible with the basic theory of other mathematical subjects, such as linear algebra and algebraic geometry.

Terry, this can be done for matrices. In fact, Harm Derksen wrote a nice proof of it in the Monthly (2003). The structure of his inductive argument actually resembles very much Gauss' approach in one of his proofs.
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Andres CaicedoSep 20 '10 at 0:50

1

Can't one prove the spectral theorem by applying the method of Lagrange multipliers to find the extremal points of some function? I seem to remember some proof along these lines...
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MarkApr 16 '11 at 12:22

and it is based on the notion of the degree of a map. Recall that for a smooth proper mapping of oriented manifolds, its degree is defined by picking a regular value and adding up the signs of the determinants of the differential of the mapping at the points in the inverse image. That the degree is well-defined is rather complicated to prove, but it explains the following topological fact.

Fact: Let $M^n$ and $N^n$ be smooth connected oriented manifolds and
$f \colon M^n \rightarrow N^n$ be a smooth proper mapping of degree
not equal to zero. Then $f$ is surjective.

To prove the Fundamental Theorem of Algebra in a "real" version, we will focus on polynomials of even degree (any of odd degree have a real root). We will show that
any real polynomial of degree $2n$ can be factored into a product of $n$ polynomials
of the second degree.

For any $d \geq 1$, identify the nonzero polynomials of degree at most $d$, considered up to scaling by nonzero real numbers, with $\mathbf P^d(\mathbf R)$ by $[a_dx^d + a_{d-1}x^{d-1} +\cdots + a_0] \mapsto
[a_d,a_{d-1},\dots,a_0]$. (The polynomials of exact degree $d$, after being scaled to be monic, are a copy of $\mathbf R^d$ in $\mathbf P^d(\mathbf R)$.) Consider the multiplication mapping
$$
\widehat{u} \colon (\mathbf P^2(\mathbf R))^n \rightarrow \mathbf P^{2n}(\mathbf R) \ \ \text{ where} \ \ \ ([f_1],[f_2],\dots,[f_n]) \mapsto [f_1f_2\cdots f_n].
$$
The mapping $\widehat{u}$ is proper since it is defined on a compact manifold and is continuous.

The mapping $\widehat{u}$ is a natural "compactification" of the mapping $u$. The space
$(\mathbf P^2(\mathbf R))^n$ can be written as the union of $(\mathbf R^{2})^n$ and an "infinitely distant part" $B_1$ ($n$-tuples of polynomials of degree at most 2 where at least one polynomial has degree less than 2), while $\mathbf P^{2n}(\mathbf R)$ can be written as the union of $\mathbf R^{2n}$ and an "infinitely distant part" $B_2$ (polynomials of degree less than $2n$). From this point of view, $\widehat{u}$ on $(\mathbf R^2)^n$ agrees with $u$ and, clearly, $\widehat{u}^{-1}(B_2) = B_1$. Therefore the map $u$ is proper.

Next we show the degree of $u$ is equal to $n!$.
Orient the space of monic polynomials of degree 2 (we denote this space as $\mathbf R^2$) arbitrarily and give $(\mathbf R^2)^n$ the product orientation (as a product of oriented manifolds). As an exercise, show
the polynomial $p(x) = \prod_{i=1}^n (x^2+i)$ is a regular value of the mapping $u$. (Hint:
This polynomial is a product of distinct monic irreducibles. Now use the description of the regular values of the multiplication mappings $\mu_k$ in Pukhlikov's proof of the Fundamental Theorem of Algebra, which is written in a separate answer on this page.)

The polynomial $p(x)$ has $n!$ inverse images under $u$: all ordered $n$-tuples
with coordinates $x^2+i$ for $i = 1,\dots,n$. Let's prove that these points
all contribute the same sign to the degree.

The mapping $u$ is invariant under permutations of its arguments, and
any such permutation preserves orientation (exercise). Therefore the
sign of the determinant of the differential at all the inverse images is the same, which shows $u$ has degree $n!$. By the topological fact at the start, $u$ is surjective, so all monic real polynomials of degree $2n$ are a product of monic quadratic real polynomials.

There is one short proof on wikipedia, that shows the statement that any endomorphism $A$ of a finite-dimensional vector space $V$ of positive dimension has an eigenvalue. Look at the resolvent function $R_A:z \mapsto (A-z)^{-1}$. Outside the disc of radius $\|A\|$, it can be developped into a geometric series. Use this geometric series to compute the integral of $R_A$ around a big circle; the result is $2 \pi i \cdot id_V$. On the other hand, if the spectrum were empty, then the resolvent were holomorphic and by Cauchy's theorem, the integral is zero. This is a contradiction if $dim (V) >0$.

To derive the FTA from the existence of eigenvalues (without using determinants), let $p(z)$ be a normed polynomial of degree $n$ and look at the vector space $V:= \mathbb{C}[z]/(p(z))$ and let $A:V \to V$ be the multiplication by $z$. $dim (V)=n$ is easy to see (division with remainder) and clearly $p(A)=0$. Take a nonzero $v \in V$ with $Av=\lambda v$. Then $0=p(A) v = p(\lambda) v$ shows that $\lambda$ is a zero of $p$.

Perhaps it could be of interest for you to know that there exists purely geometric proofs of this result. Concretely, it can be shown that, if the FTA fails then there exists a plane Riemannian metric over the Sphere S^2. Of course, this produces a contradiction since the sphere is not flat. This proof can be located at the very recent paper by J M Almira and A Romero: "Some Riemannian Geometric proofs of the fundamental theorem of algebra", which is available at: http://www.mathem.pub.ro/dgds/v14/D14-al.pdf

Thanks to Tim Chow for citing me. Technically, you don't need to show every polynomial of prime degree in F[x] has a root, you just need to show that there is a field G such that every polynomial of odd prime degree in G[x] has a root and every element or its additive inverse has a square root; then G[i] will be algebraically closed. Even more interesting, to show that all polynomials of degree d have a root, all you need is that all polynomials of degree p have a root for those p which divide d, plus the existence of any sufficiently large degree d' such that all polynomials of degree d' have a root (an explicit algorithm for how large d' must be is easily derivable from my proof).

Of course, this is not a proof of the Fundamental Theorem of Algebra, what I did was identify the pure algebraic core of the requirement that a field be algebraically closed. To show that the complex numbers are algebraically closed, you still need some way of showing that real polynomials of odd prime degree have roots, which depends on the Intermediate Value Theorem or some other analytical or topological argument in all the proofs I know.

If $-1$ has no square root in $G$, then the condition “every polynomial of odd prime degree has a root” trivially implies “every polynomial of odd degree has a root”, and this has nothing to with with primes (any infinite set of odd natural numbers will do): given an odd degree polynomial $f$, consider $(x^2+1)^df$, where $d$ is chosen so that $\deg(f)+2d$ is prime.
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Emil JeřábekJul 31 '12 at 12:34

Here's another complex analysis proof that I heard about for the first time under a week ago (because it was set as a question on a course I am teaching for). Pick a circle large enough for the modulus of p(z) to be greater than |p(0)| everywhere in that circle. Inside that circle take a point w where the modulus of p is minimal (which obviously you can do by compactness). There are many ways of proving that p(w)=0. One can use the minimum modulus theorem (that any point of minimum modulus not on the boundary must be a zero), the open mapping theorem, the local mapping theorem, or an elementary bare-hands argument.

If $p$ is a non-constant polynomial without roots then $f = \text{Re}(\frac{1}{p})$ is a bounded harmonic function which goes to zero at infinity. Consider the gradient flow for $f$. This flow is area preserving because $f$ is harmonic. Also, the value of $f$ is strictly increasing along the orbit of any non-singular point.

Consider the bounded set where $f > \epsilon > 0$. This set is invariant for the flow and by Poincarè's recurrence theorem almost all orbits in it are recurrent. However because of the monotonicity of the values of $f$ along orbits this is impossible unless all these orbits are singular. Hence the derivative of $f$ on the set where it is positive is zero. The same argument applied to $-f$ shows that the derivative of $f$ is zero on the set where $f$ is negative. Hence $f$ is constant.

The same argument applied to the imaginary part of the reciprocal of $p$ implies that $p$ is constant.

I'm partial to Milnor's proof in Topology from the Differentiable Viewpoint, a slightly simpler variant of the "every complex non-constant polynomial $p$ is surjective" proof given above, published somewhat earlier (1965). In brief:

Proof of the F.T.A.: Using stereographic projection, we can consider the polynomial as a smooth map $f: S^2 \to S^2$. Since $f$ has only a finite number of critical points, the set of regular values is connected; the locally constant $\#f^{-1}$ is therefore constant on this set. As $\#f^{-1}$ cannot be zero for all regular values of $f$, it must be zero for none. Thus $f$ is surjective, and the polynomial has a root.

I came across this article when I was pondering teaching the FTA to my multi-variable calculus class. In the end, I didn't have time to include it. It is nice in that it relies only on Green's theorem which we get through in the first semester. On the other hand, it is clear that they are largely being clever at avoiding introducing new definitions or standard theorems (for instance they use Cauchy-Riemann equations for polynomials without really saying so). I think it's nice nevertheless.

The referee assigned to review that Monthly article didn't carry out proper due diligence: that proof goes back to Gauss! See, for instance, the discussion of the third proof of Gauss in math.huji.ac.il/~ehud/MH/Gauss-HarelCain.pdf. I saw this proof for the first time in Volume II of Fikhtengoltz's "Course of Differential and Integral Calculus" and translated it into English in case I want to use it myself. It's posted on math.uconn.edu/~kconrad/blurbs/analysis/fundthmalgcalculus.pdf and at the end the hidden connection with the argument principle is indicated.
–
KConradJan 17 '10 at 19:30

4

The place where that article appeared, the American Math. Monthly, is not intended specifically for original papers. But the author really found essentially the proof of Gauss and neither he nor the referee realized it. It should have been presented in the article as a proof that goes back to Gauss but is not so well-known (except if you translate it into complex variables it becomes one of the known complex-analytic proofs). I don't have a problem with the proof appearing in the Monthly, but the lack of awareness that it is a known proof is kind of unfortunate.
–
KConradFeb 7 '10 at 1:52

When you consider how polynomial $f$ of degree $n$ acts on a big circle $R$, it gives rise to a map $S^1 \to S^1$ of degree $n$. Such a map cannot be continued to a map of a disk $D \to D-\{0\}$, thus $f(D)$ contains point 0.

which Ilya mentioned as being only in Russian so far. What I present below is not a literal translation (as if anyone on this site cares...).

The argument will use only real variables: there is no use of complex numbers anywhere.
The goal is to show for every $n \geq 1$ that each monic polynomial of degree $n$ in ${\mathbf R}[X]$ is a product of linear and quadratic polynomials.
This is clear for $n = 1$ and 2, so from now on let $n \geq 3$ and assume by induction
that nonconstant polynomials of degree less than $n$ admit
factorizations into a product of linear and quadratic polynomials.

First, some context: we're going to make use of proper mappings. A complex-variable proof on this page listed by Gian depends on the fact that a nonconstant one-variable complex polynomial is a proper mapping $\mathbf C \rightarrow \mathbf C$. Of course a nonconstant one-variable real polynomial is a proper mapping $\mathbf R \rightarrow \mathbf R$, but that is not the kind of proper mapping we will use. Instead, we will use the fact (to be explained below) that multiplication of real one-variable polynomials of a fixed degree is a proper mapping on spaces of polynomials. I suppose if you find yourself teaching a course where you want to give the students an interesting but not well-known application of the concept of a proper mapping, you could direct them to this argument.

Now let's get into the proof. It suffices to focus on monic polynomials and their monic factorizations.
For any positive integer $d$, let $P_d$ be the space of monic polynomials
of degree $d$:
$$
x^d + a_{d-1}x^{d-1} + \cdots + a_1x + a_0.
$$
By induction, every polynomial in $P_1, \dots, P_{n-1}$ is a product of linear and quadratic polynomials. We will show every polynomial in $P_n$ is a product of
a polynomial in some $P_k$ and $P_{n-k}$ where $1 \leq k \leq n-1$ and therefore
is a product of linear and quadratic polynomials.

For $n \geq 3$ and $1 \leq k \leq n-1$, define the multiplication map
$$\mu_k \colon P_k \times P_{n-k} \rightarrow P_n \ \ \text{ by } \ \ \mu_k(g,h) = gh.$$ Let $Z_k$ be the image of $\mu_k$ in $P_n$ and
$$Z = \bigcup_{k=1}^{n-1} Z_k.$$
These are the monic polynomials of degree $n$ which are composite. We want
to show $Z = P_n$. To achieve this we will look at topological properties of $\mu_k$.

We can identify $P_d$ with ${\mathbf R}^d$ by associating to the polynomial displayed way up above the vector $(a_{d-1},\dots,a_1,a_0)$. This makes $\mu_k \colon P_k \times P_{n-k} \rightarrow P_n$ a continuous mapping. The key point
is that $\mu_k$ is a proper mapping: its inverse images of compact sets are
compact. To explain why $\mu_k$ is proper, we will use an idea of Pushkar' to "compactify" $\mu_k$
to a mapping on projective spaces. (In the journal where Pukhlikov's paper appeared, the paper by Pushkar' with his nice idea comes immediately afterwards. Puklikov's own approach to proving $\mu_k$ is proper is more complicated and I will not be translating it!)

Let $Q_d$ be the nonzero real polynomials of degree $\leq d$ considered
up to scaling. There is a bijection
$Q_d \rightarrow {\mathbf P}^d({\mathbf R})$ associating to a class of polynomials
$[a_dx^d + \cdots + a_1x + a_0]$ in $Q_d$ the point $[a_d,\dots,a_1,a_0]$.
In this way we make $Q_d$ a compact Hausdorff space.
The monic polynomials $P_d$, of degree $d$, embed into
$Q_d$ in a natural way and are identified in ${\mathbf P}^d({\mathbf R})$
with a standard copy of ${\mathbf R}^d$.

Define $\widehat{\mu}_k \colon Q_k \times Q_{n-k} \rightarrow Q_n$
by $\widehat{\mu}_k([g],[h]) = [gh]$.
This is well-defined and restricts on the embedded subsets of monic polynomials to the
mapping $\mu_k \colon P_k \times P_{n-k} \rightarrow P_n$. In natural homogeneous coordinates, $\widehat{\mu}_k$ is a polynomial mapping so
it is continuous. Since projective spaces are compact and Hausdorff,
$\widehat{\mu}_k$ is a proper map. Then, since
$\widehat{\mu}_k^{-1}(P_n) = P_k \times P_{n-k}$,
restricting $\widehat{\mu}_k$ to $P_k \times P_{n-k}$ shows $\mu_k$ is proper.

Since proper mappings are closed mappings,
each $Z_k$ is a closed subset of $P_n$, so $Z = Z_1 \cup \cdots \cup Z_{n-1}$
is closed in $P_n$. Topologically, $P_n \cong {\mathbf R}^n$ is connected,
so if we could show $Z$ is also open in $P_n$ then we immediately
get $Z = P_n$ (since $Z \not= \emptyset$), which was our goal. Alas, it will not be easy to show $Z$ is open directly, but a modification of this
idea will work.

We want to show that if a polynomial $f$ is in $Z$ then all polynomials in $P_n$ that are near
$f$ are also in $Z$. The inverse function theorem is a natural tool to use in
this context: supposing $f = \mu_k(g,h)$, is the Jacobian determinant of
$\mu_k \colon P_k \times P_{n-k} \rightarrow P_n$ nonzero at $(g,h)$?
If so, then $\mu_k$ has a continuous local inverse defined in a neighborhood of $f$.

If $g$ and $h$ are relatively prime then
every polynomial of degree less than $n$ is uniquely of the form
$gv + hu$ where $\deg u < \deg g$ or $u = 0$ and $\deg v < \deg h$ or $v = 0$,
while if $g$ and $h$ are not relatively prime then
we can write $gv + hu = 0$ for some nonzero polynomials
$u$ and $v$ where $\deg u < \deg g$ and $\deg v < \deg h$.
Therefore the Jacobian of $\mu_k$ at $(g,h)$ is invertible if $g$ and $h$ are relatively prime
and not otherwise.

We conclude that if $f \in Z$ can be written somehow as a product
of nonconstant relatively prime polynomials then a neighborhood of $f$ in $P_n$ is inside $Z$.
Every $f \in Z$ is a product of linear and quadratic polynomials, so
$f$ can't be written as a product of nonconstant relatively prime
polynomials precisely when it is a power of a linear or quadratic polynomial. Let $Y$ be all these "degenerate" polynomials in $P_n$:
all $(x+a)^n$ for real $a$ if $n$ is odd and all $(x^2+bx+c)^{n/2}$ for real $b$ and $c$ if $n$ is even. (Note when $n$ is even that $(x+a)^n = (x^2 + 2ax + a^2)^{n/2}$.) We have shown $Z - Y$ is open in $P_n$. This is weaker than our hope of
showing $Z$ is open in $P_n$. But we're in good shape, as long as
we change our focus from $P_n$ to $P_n - Y$. If $n = 2$ then $Y = P_2$ and $P_2 - Y$ is empty.
For the first time we will use the fact that $n \geq 3$.

Identifying $P_n$ with ${\mathbf R}^n$ using polynomial coefficients,
$Y$ is either an algebraic curve ($n$ odd) or algebraic surface ($n$ even) sitting in
${\mathbf R}^n$. For $n \geq 3$, the complement of an algebraic curve
or algebraic surface in ${\mathbf R}^n$ for $n \geq 3$ is path connected, and thus
connected.

The set $Z-Y$ is nonempty since $(x-1)(x-2)\cdots(x-n)$ is in it. Since $Z$ is closed in $P_n$, $Z \cap (P_n - Y) = Z - Y$ is closed in $P_n - Y$.
The inverse function theorem tells us that
$Z - Y$ is open in $P_n$, so it is open in $P_n - Y$.
Therefore $Z - Y$ is a nonempty open and closed subset of $P_n - Y$.
Since $P_n - Y$ is connected and $Z - Y$ is not empty, $Z - Y = P_n - Y$.
Since $Y \subset Z$, we get $Z = P_n$ and this completes Pukhlikov's "real" proof
of the Fundamental Theorem of Algebra.